1. Rewrite the differential equation

Given $x\,dy-(y^2-4y)\,dx=0,$ for $x>0$.

Divide by $dx$: $x\frac{dy}{dx}=y^2-4y=y(y-4).$ So $\frac{dy}{dx}=\frac{y(y-4)}{x}.$

This is separable: $\frac{dy}{y(y-4)}=\frac{dx}{x}.$

2. Perform partial fraction decomposition

We write $\frac{1}{y(y-4)}=\frac{A}{y}+\frac{B}{y-4}.$ Then $1=A(y-4)+By=(A+B)y-4A.$ Comparing coefficients: $A+B=0, \qquad -4A=1.$ Thus $A=-\frac14, \qquad B=\frac14.$

Hence $\frac{1}{y(y-4)}=-\frac{1}{4y}+\frac{1}{4(y-4)}.$

So $\int \frac{dy}{y(y-4)}=\frac14\int\left(\frac{1}{y-4}-\frac{1}{y}\right)dy.$

3. Integrate

Therefore, $\frac14\left(\ln|y-4| - \ln|y|\right)=\ln x + C.$ So $\ln\left|\frac{y-4}{y}\right|=4\ln x + C_1.$

Exponentiating, $\frac{y-4}{y}=Cx^4.$

Now solve for $y$: $y-4=Cx^4y$ $y(1-Cx^4)=4$ $y=\frac{4}{1-Cx^4}.$

4. Use the initial condition $y(1)=2$

Substitute $x=1$, $y=2$: $2=\frac{4}{1-C}.$ Thus $1-C=2 \implies C=-1.$

Hence the solution is $y(x)=\frac{4}{1+x^4}.$

5. Check the slope is never zero

Differentiate from the DE: $\frac{dy}{dx}=\frac{y(y-4)}{x}.$ For our solution, $y=\frac{4}{1+x^4}.$ Since $x>0$, we have $0<y<4$ for $x>0$, and $y=0$ or $y=4$ never occurs. Hence $y(y-4)\neq 0,$ so $\frac{dy}{dx}\neq 0$ for all $x>0$. Thus the given condition is satisfied.

6. Find $y(\sqrt{2})$

Since $(\sqrt{2})^4=4,$ we get $y(\sqrt{2})=\frac{4}{1+4}=\frac45.$ Therefore, $10y(\sqrt{2})=10\cdot \frac45=8.$

7. Comparison with stored answer

Our derived answer is $8$, which matches the stored correct answer.